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Jul 30, 2020 at 19:11 vote accept vidyarthi
Jul 30, 2020 at 18:55 comment added Gregory J. Puleo I suspect there would still be far too many previously-colored adjacent edges for the late edges within $A$. Note that the last edge considered within $A$ will have $2p-1$ previously-colored adjacent edges just within $A$, and therefore couldn't afford to be incident to any edges going to $B$. I think the only way to relax the hypothesis you stated in the question and have this proof still go through is to allow vertex $a_i$ to have degree at most $i-1$ in $B$, rather than degree exactly $i-1$ (and likewise for $B$-vertices).
Jul 30, 2020 at 18:52 comment added vidyarthi ok, let us limit the degree of the bipartite graph to a maximum of $p$, then I think it should be possible,right?
Jul 30, 2020 at 18:46 comment added Gregory J. Puleo The degree constraint is essential here for arguing that each edge has few enough previously-colored adjacent edges going to $B$. In the extreme case where the bipartite graph was $K_{p+1, p+1}$, the whole graph would just be $K_{2p+2}$, and then no matter how you slice it the last edge you consider will have $2(2p+1) - 1 = 4p+1$ previously-colored adjacent edges.
Jul 30, 2020 at 18:03 comment added vidyarthi by the way, would replacing the bipartite graph with arbitrary bipartite graph have any effect (I dont think so)? If so, then I think we could extend this method to prove edge chromatic choosability for all graphs with maximum degree$\ge\frac{n}{2}$, whre $n$ is the order of the graph
Jul 30, 2020 at 18:00 comment added vidyarthi great! but I think first coloring the edges of $A$ (or $B$) and then the edges of the bipartite graph and lastly $B$ (or $A$) would also work. But, for this, I would use the paer refereed and the Galvin' stheorem.
Jul 29, 2020 at 22:22 history edited Gregory J. Puleo CC BY-SA 4.0
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Jul 29, 2020 at 21:29 history answered Gregory J. Puleo CC BY-SA 4.0